Question

import java.util.*;  
import java.applet.*; 
import java.util.Scanner; 
public class Game
{
public static void main(String[] args){  

System.out.println("I am ready to play!");
  Scanner input = new Scanner(System.in);
        System.out.println("What's your age?");
        String userinput = input.nextLine();
        if (userinput < 13){
            System.out.println("You are allowed to play at your own risk");
        } else
            System.out.println("Play On!");
        }
System.out.println("You are at an Iggy concert, and you hear this lyric 'Are you ready?, start       running.'");
System.out.println("Suddenly, Iggy stops and says, 'Who wants to race me at running?'");

  Scanner input = new Scanner(System.in);
        System.out.println("Do you want to race Iggy on stage?");
        String userinput = input.nextLine();
        if (userinput == yes){
            "You and Iggy start racing. It's neck and neck! You win by a shoelace!");
        } else
            System.out.println("Oh no! Iggy shakes his head and sings 'I set a pace, so I can race without pacing.'");

var feedback = prompt("Rate your game out of 10");
if(feedback > 8)
{
    console.log("Thank you! You should race again at the next concert!");
}
else
{
    console.log("I'll keep practicing coding and racing.");
}
}
}

这是高中Java 1课程的作业，我不知道如何修复所有这些。我意识到有很多错误，除了我之外，这对任何人来说都是无关紧要的，但我想真正学习java，这个课程并没有教我任何东西，所以任何帮助都会非常感激。

Answer 1

首先开始阅读java.util.Scanner＆amp;然后阅读字符串比较！我不会给你答案，但提示！

有些错误是：
1.你需要使用.equals而不是==

String userinput = input.nextLine();
    if (userinput == yes)

你需要一个int而不是在Interger.parseInt（String）上读取的字符串

String userinput = input.nextLine（）;
if（userinput＆lt; 13）{}

什么是var，什么是提示？

var feedback = prompt（“为你的游戏评分10分”）;

console is not defined

console.log（“谢谢！你应该在下一场音乐会上再次参加比赛！”）;

这是两次定义。

扫描仪输入=新扫描仪（System.in）;

＆安培;等等!!!

Java 1的调试练习

1 个答案: